Đề bài sai nhé, từ giả thiết chỉ xác định được \(x+y=0\Rightarrow y=-x\)

\(\Rightarrow A=4x^2-x^2+x^2+15=4x^2+15\) ko rút gọn được

a/ Vế trái = x³ + x² + x - x² - x - 1 = x³ - 1 (= vế phải)

b/ hình như đề sai

b/ Vế trái = x⁴ + x³y + x²y² + xy³ - x³y - x²y² - xy³ - y⁴ = x⁴ - y⁴ (= vế phải)

b)(x^3+x^2y+xy^2+y^3)(x-y)=x^4-y^4

xy + 2x + y + 2 = y(x + 1) + 2(x + 1) = (x + 1).(y + 2)

x(x - 1) + x(x + 3) = x(x - 1 + x + 3) = x. ( 2x + 2) = 2x.(x + 1)

\(-4x^2+8x-4=-4\left(x^2-2x+1\right)=-4\left(x-1\right)^2\)

c: \(-4x^2+8x-4\)

\(=-4\left(x^2-2x+1\right)\)

\(=-4\left(x-1\right)^2\)

a) \(xy+2x+y+2\)

\(=\left(xy+y\right)+\left(2x+2\right)\)

\(=y\left(x+1\right)+2\left(x+1\right)\)

\(=\left(y+2\right)\left(x+1\right)\)

b) \(x\left(x-1\right)+x\left(x+3\right)\)

\(=x^2-x+x^2+3x\)

\(=2x^2+2x\)

\(=2x\left(x+1\right)\)

c) \(-4x^2+8x-4\)

\(=-\left[\left(2x\right)^2-2.2x.2+2^2\right]\)

\(=-\left(2x-2\right)^2\)

\(\left(x+2\right)\left(x^2+2x-9\right)\)

\(=x^3+2x^2-9x+2x^2+4x-18\)

\(=x^3+4x^2-5x-18\)

\(\left(x^{2y}-6\right)\left(x^2-5\right)\)

\(=x^{4y}-5x^{2y}-6x^2+30\)

\(\left(x+y\right)\left(xy-4+y\right)\)

\(=x^2y-4x+xy+xy^2-4y+y^2\)

câu còn lại tương tự  nha

Lời giải:

$\frac{1}{x}+\frac{1}{y}+\frac{2}{x+y}=\frac{x+y}{xy}+\frac{2}{x+y}$

$=x+y+\frac{2}{x+y}$

$=\frac{x+y}{2}+\frac{x+y}{2}+\frac{2}{x+y}$

$\geq \frac{x+y}{2}+2\sqrt{\frac{x+y}{2}.\frac{2}{x+y}}$ (áp dụng BDT Cô-si)

$\geq \frac{2\sqrt{xy}}{2}+2=\frac{2}{2}+2=3$

Vậy ta có đpcm

Dấu "=" xảy ra khi $x=y=1$

tích đi rồi t làm 

9 T I C H  sai buồn

\(A=\frac{\sqrt{x^3}}{\sqrt{xy}-2y}-\frac{2x}{x+\sqrt{x}-2\sqrt{xy}-2\sqrt{y}}.\frac{1-x}{1-\sqrt{x}}..\)

nhờ vào năng lực rinegan tối hậu của ta , ta có thể dễ dàng nhìn thấy mẫu chung 

\(x+\sqrt{x}-2\sqrt{xy}-2\sqrt{y}=\sqrt{x}\left(\sqrt{x}-2\sqrt{xy}\right)+\left(\sqrt{x}-2\sqrt{y}\right)=\left(\sqrt{x}-2\sqrt{y}\right)\left(\sqrt{x}+1\right)\)

\(A=\frac{\sqrt{x^3}}{\sqrt{y}\left(\sqrt{x}-2\sqrt{y}\right)}-\frac{2x\left(x-1\right)}{\left(\sqrt{x}-2\sqrt{y}\right)\left(1+\sqrt{x}\right)\left(1-\sqrt{x}\right)}.\)

\(\left(1+\sqrt{x}\right)\left(1-\sqrt{x}\right)=1-x\)

\(A=\frac{\sqrt{x^3}-2x\sqrt{y}}{\sqrt{y}\left(\sqrt{x}-2\sqrt{y}\right)}=\frac{x\sqrt{x}-2x\sqrt{y}}{\sqrt{y}\left(\sqrt{x}-2\sqrt{y}\right)}=\frac{x\left(\sqrt{x}-2\sqrt{y}\right)}{\sqrt{y}\left(\sqrt{x}-2\sqrt{y}\right)}=\frac{x}{\sqrt{y}}\)

b) thay y=625 vào ta được

\(\frac{x}{\sqrt{625}}=\frac{x}{25}< 0.2\Leftrightarrow x< 5\)

vậy   \(0< x< 5\)

a) \(\left(x+2y\right)^2=x^2+2.x.2y+\left(2y\right)^2=x^2+4xy+4y^2\)

b) \(\left(3-x\right).\left(3+x\right)=9+3x-3x-x^2=9-x^2=3^2-x^2\)

c) \(\left(5-x\right)^2=5^2-2.5.x+x^2=25-10x+x^2\)

d) \(\left(3+y\right)^2=3^2+2.3.y+y^2=9+6y+y^2\)

a) x²+4xy+4y² b)x(9-x²) = 9x-x³ c)25-10x+x² d)9+6y+x²

​

Bài làm:

a) \(\left|\frac{1}{2}x-\frac{5}{2}\right|-1=-\frac{1}{2}\)

\(\Leftrightarrow\left|\frac{1}{2}x-\frac{5}{2}\right|=\frac{1}{2}\)

\(\Leftrightarrow\orbr{\begin{cases}\frac{1}{2}x-\frac{5}{2}=\frac{1}{2}\\\frac{1}{2}x-\frac{5}{2}=-\frac{1}{2}\end{cases}}\Leftrightarrow\orbr{\begin{cases}\frac{1}{2}x=3\\\frac{1}{2}x=2\end{cases}}\Rightarrow\orbr{\begin{cases}x=6\\x=4\end{cases}}\)

+ Nếu x = 6

\(\left|12-\frac{1}{3}y\right|=\frac{5}{6}\)

\(\Leftrightarrow\orbr{\begin{cases}12-\frac{1}{3}y=\frac{5}{6}\\12-\frac{1}{3}y=-\frac{5}{6}\end{cases}}\Leftrightarrow\orbr{\begin{cases}\frac{1}{3}y=\frac{67}{6}\\\frac{1}{3}y=\frac{77}{6}\end{cases}}\Rightarrow\orbr{\begin{cases}y=\frac{67}{2}\\y=\frac{77}{2}\end{cases}}\)

+ Nếu x = 4

\(\left|8-\frac{1}{3}y\right|=\frac{5}{6}\)

\(\Leftrightarrow\orbr{\begin{cases}8-\frac{1}{3}y=\frac{5}{6}\\8-\frac{1}{3}y=-\frac{5}{6}\end{cases}}\Leftrightarrow\orbr{\begin{cases}\frac{1}{3}y=\frac{43}{6}\\\frac{1}{3}y=\frac{53}{6}\end{cases}}\Rightarrow\orbr{\begin{cases}y=\frac{43}{2}\\y=\frac{53}{2}\end{cases}}\)

Vậy ta có 4 cặp số (x;y) thỏa mãn: \(\left(6;\frac{67}{2}\right);\left(6;\frac{77}{2}\right);\left(4;\frac{43}{2}\right);\left(4;\frac{53}{2}\right)\)

b) \(\frac{3}{2}x-\frac{1}{2}\left(x-\frac{2}{3}\right)=\frac{5}{3}\)

\(\Leftrightarrow\frac{3}{2}x-\frac{1}{2}x+\frac{1}{3}=\frac{5}{3}\)

\(\Leftrightarrow x=\frac{4}{3}\)

Thay vào ta được:

\(\frac{2.\frac{4}{3}+y}{\frac{4}{3}-2y}=\frac{5}{4}\)

\(\Leftrightarrow\frac{32}{3}+4y=\frac{20}{3}-10y\)

\(\Leftrightarrow14y=-4\)

\(\Rightarrow y=-\frac{2}{7}\)

Vậy ta có 1 cặp số (x;y) thỏa mãn: \(\left(\frac{4}{3};-\frac{2}{7}\right)\)